If the two particles are indistinguishable, the Hilbert space is the antisymmetrized tensor product of two 1-particle spaces, both before and after measurement.
If you want to only consider the spins, then the 1-particle Hilbert space is 2 dimensional : $\mathfrak h = \mathbb C^2$. Then, there is only one 2-particle state, which is :
$$\frac{1}{\sqrt 2}\Big(|1\rangle_a|0\rangle_b - |0\rangle_a|1\rangle_b\Big) $$
Then you cannot actually measure the 'first' particle, because they are indistinguishable. The measurement gives you "one of the two particles is in the state $|1\rangle$, and the state remains unchanged.
If you want to say, the particle $a$ is in the state $|1\rangle$", then you need both particles to be distinguishable, in which case the Hilbert space is the full tensor product both before and after measurement.
If you want to describe the situation where you have two particles in two different boxes, with two possible internal states, then you need a $4$-dimensional $1$-particle state. For example, let the state $|A1\rangle$ be "the particle is in the box $A$ with internal state $1$ and the other states of the basis be $|A0\rangle,|B1\rangle,|B0\rangle$.
Then, the initial state "there is a particle in each box and their internal states are different" is :
$$|i\rangle = \frac{1}{2}\Big(|A1\rangle|B0\rangle - |B0\rangle |A1\rangle \Big) + \frac{e^{i\varphi}}{2}\Big(|A0\rangle|B1\rangle - |B1\rangle|A0\rangle\Big)$$
"Is there a particle in the box $A$ with state $1$" is a valid measurement. The associated operator is the projector :
$$\hat P = |A1\rangle\langle A1|\otimes\mathbf 1 +\mathbf 1\otimes|A1\rangle\langle A1|$$
If you measure $|i\rangle$ and find a particle in state $1$ and in box $A$, the state after measurement is :
$$|f\rangle = \frac{1}{\sqrt 2}\Big(|A1\rangle|B0\rangle - |B0\rangle|A1\rangle\Big)$$
Here, you see that both states are antisymmetric, as is required for indistinguishable fermions. But you can also notice that the presence of the degrees of freedom given by the two boxes allow us to distinguish the two particles. Hence, you could safely take them to be distinguishable. This is often done in discussions of entanglement.